Theorem disji 5093

Description: Property of a disjoint collection: if 𝐵(𝑋) = 𝐶 and 𝐵(𝑌) = 𝐷 have a common element 𝑍, then 𝑋 = 𝑌. (Contributed by Mario Carneiro, 14-Nov-2016.)

Hypotheses

Ref	Expression
disji.1	⊢ (𝑥 = 𝑋 → 𝐵 = 𝐶)
disji.2	⊢ (𝑥 = 𝑌 → 𝐵 = 𝐷)

Assertion

Ref	Expression
disji	⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝑍 ∈ 𝐶 ∧ 𝑍 ∈ 𝐷)) → 𝑋 = 𝑌)

Distinct variable groups:   𝑥,𝐴   𝑥,𝐶   𝑥,𝐷   𝑥,𝑋   𝑥,𝑌

Allowed substitution hints:   𝐵(𝑥)   𝑍(𝑥)

Proof of Theorem disji

Step	Hyp	Ref	Expression
1		inelcm 4424	. 2 ⊢ ((𝑍 ∈ 𝐶 ∧ 𝑍 ∈ 𝐷) → (𝐶 ∩ 𝐷) ≠ ∅)
2		disji.1	. . . . . 6 ⊢ (𝑥 = 𝑋 → 𝐵 = 𝐶)
3		disji.2	. . . . . 6 ⊢ (𝑥 = 𝑌 → 𝐵 = 𝐷)
4	2, 3	disji2 5092	. . . . 5 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ 𝑋 ≠ 𝑌) → (𝐶 ∩ 𝐷) = ∅)
5	4	3expia 1138	. . . 4 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → (𝑋 ≠ 𝑌 → (𝐶 ∩ 𝐷) = ∅))
6	5	necon1d 2979	. . 3 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴)) → ((𝐶 ∩ 𝐷) ≠ ∅ → 𝑋 = 𝑌))
7	6	3impia 1134	. 2 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝐶 ∩ 𝐷) ≠ ∅) → 𝑋 = 𝑌)
8	1, 7	syl3an3 1182	1 ⊢ ((Disj 𝑥 ∈ 𝐴 𝐵 ∧ (𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴) ∧ (𝑍 ∈ 𝐶 ∧ 𝑍 ∈ 𝐷)) → 𝑋 = 𝑌)

This proof depends on syntax axioms:   → wi 4   ∧ wa 400   ∧ w3a 1102   = wceq 1569   ∈ wcel 2142   ≠ wne 2957   ∩ cin 3903  ∅c0 4285  Disj wdisj 5075

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-10 2175  ax-11 2191  ax-12 2212  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-fal 1582  df-ex 1809  df-nf 1813  df-sb 2096  df-mo 2566  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ne 2958  df-ral 3079  df-rmo 3368  df-rab 3416  df-v 3456  df-sbc 3744  df-csb 3853  df-dif 3907  df-in 3911  df-nul 4286  df-disj 5076

This theorem is used by:  volfiniun  25717  fnpreimac  33026